Comoving redshift and area of sphere

[bowlbase]

Homework Statement

Standard candles may be used to measure the "luminosity distance", using D_L =
(L/4F)^1/2, where L is the source's intrinsic luminosity, and F is the observed
flux. Inthis problem you will relate the luminosity distance to the previously discussed angular
diameter distance D_A. Consider a source at redshift z, which emits light isotropically. In its rest frame, suppose it emits a pulse of total energy E_e over a short time interval
Δt_e. Suppose that today, at z = 0, we have a telescope with collecting area
ΔA. For simplicity, assume that the universe is spatially at, with $\Omega$_curv = 0, so that the angular diameter distance is related to the comoving distance by D_A = aD_co.

The pulse of light isotropically emitted at redshift z is now, at z = 0, spread out
over the surface of a sphere. Assuming that this source is at comoving distance
D_co from us, what is the comoving area of that sphere today? What is the
proper physical area of that sphere today? Assuming that there is no intervening
absorption or scattering, what fraction of the photons emitted by the source
impinge on our telescope with collecting area Δ
A? Express your answer in terms
of D_co.

Homework Equations

The Attempt at a Solution

I think that I need to do an integral somewhere with z going from z to 0. However, at first I just thought that the the surface area (SA) would just be $4\pi(\frac{aD_{co}}{2})^2$ but D_co is just suppose to be the distance from me, the observer, to the source of the emitted energy.

Anyway, I know that SA=$4 \pi r^2$. I just need to find 'r' from z → 0. I know that $D_{co}=\int\frac{c dt}{a(t)}$. a has a time dependence so I can't just replace it with a^-1=1+z. If I do I just get $\int_z^0c(1+z)dt$. Which would be $-cz(1+z)$
I don't believe a negative distance would properly describe this situation.

Also, the second question I bolded has me confused. Isn't comoving distances suppose to describe how things actually are? How are the two bolded questions different at all?

 

[mark726]

Hello! Thank you for your post. I will try my best to help you with this problem.

Firstly, you are correct in thinking that an integral is needed to find the surface area of the sphere. However, the integral should go from z (the source's redshift) to infinity, not to 0. This is because we want to find the comoving distance of the source, which is the distance it would be if there was no expansion of the universe. At z = infinity, the source would be at its maximum comoving distance.

To find the comoving distance Dco, we can use the formula you provided: $D_{co}=\int\frac{c dt}{a(t)}$. However, since we are assuming a flat universe, we can simplify this to $D_{co}=\frac{c}{H_0}\int_z^{\infty}\frac{dz'}{E(z')}$, where E(z) = (Ωm(1+z)^3 + ΩΛ)^1/2 is the expansion parameter and H0 is the Hubble constant.

Now, to find the surface area of the sphere, we can use the formula SA = $4 \pi r^2$, where r is the comoving distance. So, the surface area at z = infinity would be $4 \pi D_{co}^2$. However, we are interested in the surface area at z = 0, so we need to take into account the expansion of the universe. This can be done by using the scale factor a(t) = (1+z)^-1. So, the surface area at z = 0 would be $4 \pi a^2(t)D_{co}^2$.

For the second question, the comoving distance Dco is indeed the distance from the observer to the source. However, the physical area is the actual size of the sphere in physical units (e.g. meters). So, the proper physical area would be $4 \pi D_{co}^2$.

Finally, to find the fraction of photons that impinge on our telescope, we can use the formula F = L/4πDl^2, where L is the source's intrinsic luminosity and Dl is the luminosity distance. We can rearrange this to get Dl = (L/4πF)1/2. Sub